To find the area of the shaded region bounded by the curves $x = y^2 - 2$ and $x = e^y$ between $y = 1$ and $y = -1$, we need to calculate the integral of the difference between these two functions over the given interval.

Step-by-Step Solution:

1. Find the points of intersection: To find the points of intersection between the curves $x = y^2 - 2$ and $x = e^y$, we need to set them equal to each other:

   $y^2 - 2 = e^y$

   This equation does not have a simple algebraic solution, so let's check the values of $y$ in the interval $[-1, 1]$:

   • For $y = -1$:
     $(-1)^2 - 2 = 1 - 2 = -1, \quad e^{-1} \approx 0.3679$

     The curves do not intersect here.

   • For $y = 0$:
     $0^2 - 2 = -2, \quad e^0 = 1$

     No intersection.

   • For $y = 1$:
     $1^2 - 2 = -1, \quad e^1 \approx 2.718$

     No intersection.

   Thus, the curves do not intersect between $y = -1$ and $y = 1$.

2. Determine which function is on top: We need to determine which function is greater on the interval $[-1, 1]$. Evaluating at points within this interval:

   • For $y = -1$:
     $y^2 - 2 = -1$, $e^y \approx 0.3679$. Here, $e^y > y^2 - 2$.

   • For $y = 0$:
     $y^2 - 2 = -2$, $e^y = 1$. Again, $e^y > y^2 - 2$.

   • For $y = 1$:
     $y^2 - 2 = -1$, $e^y \approx 2.718$. Again, $e^y > y^2 - 2$.

   Therefore, $e^y$ is the upper function and $y^2 - 2$ is the lower function over this interval.

3. Set up the integral to find the area: The area $A$ between the curves from $y = -1$ to $y = 1$ is given by:

   $A = \int_{-1}^{1} \left(e^y - (y^2 - 2)\right) \, dy$

   Simplify the integrand:

   $A = \int_{-1}^{1} \left(e^y - y^2 + 2\right) \, dy$

4. Compute the integral:

   The integral can be split into three parts:

   $A = \int_{-1}^{1} e^y \, dy - \int_{-1}^{1} y^2 \, dy + \int_{-1}^{1} 2 \, dy$

   Let's compute these integrals:

   • $\int e^y \, dy = e^y$, so:

     $\left[ e^y \right]_{-1}^{1} = e^1 - e^{-1} = e - \frac{1}{e}$

   • $\int y^2 \, dy = \frac{y^3}{3}$, so:

     $\left[ \frac{y^3}{3} \right]_{-1}^{1} = \frac{1^3}{3} - \frac{(-1)^3}{3} = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3}$

   • $\int 2 \, dy = 2y$, so:

     $\left[ 2y \right]_{-1}^{1} = 2(1) - 2(-1) = 2 + 2 = 4$

5. Combine the results:

   $A = \left(e - \frac{1}{e}\right) - \frac{2}{3} + 4$

   Simplify further:

   $A = e - \frac{1}{e} + \frac{10}{3}$

Final Answer:

$\boxed{A = e - \frac{1}{e} + \frac{10}{3}}$

Would you like any further details or explanations?

Related Questions:

1. How would the area change if the interval was between $y = 0$ and $y = 2$?
2. What is the volume of the solid formed when the shaded region is rotated about the x-axis?
3. What are the points of intersection between $x = y^2 - 2$ and $x = e^y$?
4. How do you find the area between two curves if both are functions of $x$?
5. Can you find the area between the curves $y = \ln(x)$ and $y = x^2 - 4$ from $x = 1$ to $x = 3$?

Tip:

When finding the area between two curves, always check which function is on top within the given interval to ensure you subtract correctly!